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(C) To determine the reaction mechanism,we analyze the rate law based on the experimental data.Let the rate law be $	ext{Rate} = k[S]^x[Nu]^y$.From e

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$S_{N}2$

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(C) To determine the reaction mechanism,we analyze the rate law based on the experimental data.Let the rate law be $	ext{Rate} = k[S]^x[Nu]^y$.From experiment $1$ and $2$,when $[S]$ is doubled and $[Nu]$ is constant,the rate doubles ($2.2 	imes 10^{-3}$ to $4.4 	imes 10^{-3}$),so $x = 1$.From experiment $1$ and $3$,when $[Nu]$ is doubled and $[S]$ is constant,the rate doubles ($2.2 	imes 10^{-3}$ to $4.4 	imes 10^{-3}$),so $y = 1$.The overall rate law is $	ext{Rate} = k[S]^1[Nu]^1$.Since the reaction is first order with respect to both the substrate and the nucleophile,the total order is $2$.This indicates a bimolecular nucleophilic substitution reaction,which is the $S_{N}2$ mechanism.

Experiment	$[S]$	$[Nu]$	Rate
$1$	$0.1$	$0.1$	$2.2 \times 10^{-3}$
$2$	$0.2$	$0.1$	$4.4 \times 10^{-3}$
$3$	$0.1$	$0.2$	$4.4 \times 10^{-3}$

$Exp.$	$[A]_0$	$[B_2]_0$	$Rate \ (mol \ L^{-1} \ s^{-1})$
$(1)$	$0.50$	$0.50$	$1.6 \times 10^{-4}$
$(2)$	$0.50$	$1.00$	$3.2 \times 10^{-4}$
$(3)$	$1.00$	$1.00$	$3.2 \times 10^{-4}$

The following results are obtained for the reaction $S + Nu \rightarrow \text{product}$. By which reaction mechanism does this reaction occur?

Experiment $[S]$ $[Nu]$ Rate

$1$ $0.1$ $0.1$ $2.2 \times 10^{-3}$

$2$ $0.2$ $0.1$ $4.4 \times 10^{-3}$

$3$ $0.1$ $0.2$ $4.4 \times 10^{-3}$

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The following results are obtained for the reaction $S + Nu \rightarrow \text{product}$. By which reaction mechanism does this reaction occur? Experiment $[S]$ $[Nu]$ Rate $1$ $0.1$ $0.1$ $2.2 \times 10^{-3}$ $2$ $0.2$ $0.1$ $4.4 \times 10^{-3}$ $3$ $0.1$ $0.2$ $4.4 \times 10^{-3}$